91Ó°ÊÓ

Consider the formula for the angular distribution of the power as:

$\frac{\mathbf{dP}}{\mathbf{»åÎ\copyright }}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{16}{\mathbf{Ï€}}^{\mathbf{2}}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{{\left|\hat{\mathbf{r}}\mathbf{×}\left(\mathbf{u}\mathbf{×}\mathbf{a}\right)\right|}^{\mathbf{2}}}{{\left(\hat{\mathbf{r}}·\mathbf{u}\right)}^{\mathbf{5}}}$

Here, the terms are:

$$\begin{gathered} \mathbf{u}=x\hat{r}-v\hat{z} \\ \mathbf{a}=a\hat{x} \\ \hat{\mathbf{r}}=\sin \mathrm{θ}\cos \mathrm{ϕ}\hat{x}+\mathrm{sinsin}\mathrm{ϕ}\hat{y}+\cos \mathrm{θ}\hat{z}\backslash \end{gathered}$$

Determine the terms in the second fraction as:

$$\begin{gathered} d\hat{r}·u=c-v\cos \mathrm{θ} \\ =c\left(1-\mathrm{β}\cos \mathrm{θ}\right) \end{gathered}$$ …. (1)

Solve further as:

$$\begin{gathered} {\left|\hat{r}×\left(u×a\right)\right|}^2={\left|u\left(\hat{r}·a\right)-a\left(\hat{r}·u\right)\right|}^2 \\ {\left|\hat{r}×\left(u×a\right)\right|}^2={\left|ua\sin \mathrm{θ}\cos \mathrm{ϕ}-ac\left(1-\mathrm{β}\cos \mathrm{θ}\right)\hat{x}\right|}^2 \\ {\left|\hat{r}×\left(u×a\right)\right|}^2=a^2{c}^2{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^2-\left(1-\mathrm{β}\right){\sin }^2\mathrm{θ}{\cos }^2\mathrm{ϕ} \end{gathered}$$ …… (2)

From equation (1) and (2) determine the formula for the angular power as:

$$\begin{gathered} \frac{dP}{d\mathrm{Î\copyright }}=\frac{q^2}{16{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0}\frac{a^2{c}^2\left[{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^2-\left(1-{\mathrm{β}}^2\right){\sin }^2\mathrm{θ}{\cos }^2\mathrm{Ï•}\right]}{c^3{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^5} \\ \frac{dP}{d\mathrm{Î\copyright }}=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{Ï€}}^{2}c}\frac{\left[{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^2-\left(1-{\mathrm{β}}^2\right){\sin }^2\mathrm{θ}{\cos }^2\mathrm{Ï•}\right]}{{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^5}\left\{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0=\frac{1}{c^2}\right\} \end{gathered}$$

In order to determine the total power integrate over the space angle

$$\begin{gathered} d\mathrm{Î\copyright }=d\mathrm{θ}\sin \mathrm{θ}\mathrm{Ï•}asfollows \\ P=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{Ï€}}^{2}c}{∫}_0^{2\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}\frac{\left[{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^2-\left(1-{\mathrm{β}}^2\right){\sin }^2\mathrm{θ}{\cos }^2\mathrm{Ï•}\right]}{{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^5}\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•} \end{gathered}$$

Using ${∫}_0^{2\mathrm{π}}{\cos }^2\mathrm{ϕ}=\mathrm{π}$integrate over as follows:

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{π}}^{2}c}{∫}_0^{\mathrm{π}}\frac{\left[2{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^2-\left(1-{\mathrm{β}}^2\right){\sin }^2\mathrm{θ}\right]}{{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^5}\sin \mathrm{θ}d\mathrm{θ} \\ =\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{π}}^{2}c}\left(J\right) \end{gathered}$$

Solve for the integral J as follows:

$$\begin{gathered} J={∫}_{1-\mathrm{β}}^{1+\mathrm{β}}\frac{\left[2u^2-\left(1-{\mathrm{β}}^2\right)\left(1-\left(1-2u+\frac{u^2}{{\mathrm{β}}^2}\right)\right)\right]}{{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^5}\sin \mathrm{θ}d\mathrm{θ}\left\{1-\mathrm{β}\cos \mathrm{θ}=u,{\sin }^2\mathrm{θ}=1-\frac{1-2u+u^2}{{\mathrm{β}}^2},\backslash \mathrm{hfill} \\ du=\mathrm{β}\sin \mathrm{θ}d\mathrm{θ}\right\} \\ =\frac{1}{{\mathrm{β}}^3}{\left[-{\left(1-{\mathrm{β}}^2\right)}^2\frac{1}{4u^4}+2\left(1-{\mathrm{β}}^2\right)\frac{1}{3u^3}-\left(1+{\mathrm{β}}^2\right)\frac{1}{2u^2}\right]}_{1+\mathrm{β}}^{1+\mathrm{β}} \end{gathered}$$

Solve for the terms in the bracket as:

First term:

$$\begin{gathered} {\left.-\left(1+{\mathrm{β}}^2\right)\frac{1}{2u^2}\right|}_{1+\mathrm{β}}^{1+\mathrm{β}}=-\frac{1+{\mathrm{β}}^2}{2}\frac{-4\mathrm{β}}{{\left(1-\mathrm{β}\right)}^2{\left(1+\mathrm{β}\right)}^2} \\ =\frac{2\mathrm{β}{\left(1+\mathrm{β}\right)}^2}{{\left(1-{\mathrm{β}}^2\right)}^2} \end{gathered}$$

Second term:

$$\begin{gathered} {\left.2\left(1-{\mathrm{β}}^2\right)\frac{1}{3u^3}\right|}_{1+\mathrm{β}}^{1+\mathrm{β}}=\frac{2}{3}\left(1-{\mathrm{β}}^2\right)\frac{-6\mathrm{β}-2{\mathrm{β}}^3}{{\left(1-\mathrm{β}\right)}^3{\left(1+\mathrm{β}\right)}^3} \\ =\frac{4\mathrm{β}{\left(3+\mathrm{β}\right)}^2}{3{\left(1-{\mathrm{β}}^2\right)}^2} \end{gathered}$$

Third term:

$$\begin{gathered} {\left.-\left(1-{\mathrm{β}}^2\right)\frac{1}{4u^4}\right|}_{1+\mathrm{β}}^{1+\mathrm{β}}=\frac{\left(1-{\mathrm{β}}^2\right)}{4}\frac{8\mathrm{β}+8{\mathrm{β}}^3}{{\left(1-\mathrm{β}\right)}^4{\left(1+\mathrm{β}\right)}^4} \\ =\frac{2\mathrm{β}\left(1+{\mathrm{β}}^2\right)\mathrm{β}}{{\left(1-{\mathrm{β}}^2\right)}^2} \end{gathered}$$

Rewrite the integral as:

$$\begin{gathered} J=\frac{1}{3{\left(1-{\mathrm{β}}^2\right)}^2}\left[2\left(1+{\mathrm{β}}^2\right)\mathrm{β}-\frac{4}{3}\mathrm{β}\left(3+{\mathrm{β}}^2+2\mathrm{β}\left(2+{\mathrm{β}}^2\right)\right)\right] \\ =\frac{8}{3{\left(1-{\mathrm{β}}^2\right)}^2} \\ =\frac{8}{3}{\mathrm{γ}}^4\left\{\mathrm{γ}={\left(1-{\mathrm{β}}^2\right)}^{\frac{-1}{2}}\right\} \end{gathered}$$

Therefore, the expression for the power is written as:

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16\mathrm{π}c}\left(\frac{8}{3}{\mathrm{γ}}^4\right) \\ =\frac{{\mathrm{μ}}_0{q}^2{a}^2}{6\mathrm{π}c}{\mathrm{γ}}^4 \end{gathered}$$ ….. (3)

From the Lienard formula solve as:

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2{\mathrm{γ}}^6}{6\mathrm{π}c}\left(a^2-{\left|\frac{v×a}{c}\right|}^2\right) \\ =\frac{{\mathrm{μ}}_0{q}^2{\mathrm{γ}}^6}{6\mathrm{π}c}\left(a^2-a^2{\mathrm{β}}^2\right)\left\{\left|v×a\right|=va\right\} \\ =\frac{{\mathrm{μ}}_0{q}^2{a}^2}{6\mathrm{π}c}{\mathrm{γ}}^4 \end{gathered}$$$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2{\mathrm{γ}}^6}{6\mathrm{π}c}\left(a^2-{\left|\frac{v×a}{c}\right|}^2\right) \\ =\frac{{\mathrm{μ}}_0{q}^2{\mathrm{γ}}^6}{6\mathrm{π}c}\left(a^2-a^2{\mathrm{β}}^2\right)\left\{\left|v×a\right|=va\right\} \\ =\frac{{\mathrm{μ}}_0{q}^2{a}^2}{6\mathrm{π}c}{\mathrm{γ}}^4 \end{gathered}$$ …… (4)

Since, the velocity and acceleration are perpendicular to each other so the equation (3) and (4) agrees:

Therefore, the Lienard formula is

$\frac{dP}{d\mathrm{Î\copyright }}=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{Ï€}}^{2}c}\frac{\left[{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^2-\left(1-{\mathrm{β}}^2\right){\sin }^2\mathrm{θ}{\cos }^2\mathrm{Ï•}\right]}{{\left(1-\mathrm{β}\cos \mathrm{θ}\right)}^5}$

and the power found by integrating and the distribution agree to each other.